Tag: odd and even numbers

Questions Related to odd and even numbers

Let S be a set of all even integers. If the operations:
1. addition 2. subtraction 3. multiplication 4. division
are applied to any pair of numbers from S, then for which operations is the resulting number is S?

maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

  1. $1, 2, 3$ and $4$

  2. $1, 2$ and $3$ only

  3. $1$ and $3$ only

  4. $2$ and $4$ only

Show answer
Correct Option: B
Explanation:

Addition of two even numbers; subtraction of two even numbers and product of two even numbers is an even number.

Multiplication of one odd and one even integer is always :

maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

  1. Even

  2. Odd

  3. Can't be determined

  4. None of the above

Show answer
Correct Option: A
Explanation:

Even integer is an integer having unit digit as a multiple of $2$

So on multiplication with odd integer, unit digit will still remain a multiple of $2$, hence, multiplication of odd and even integer gives even integer.

$a, b, c$ are even numbers and $x, y, z$ are odd numbers. Which of the following relationships can't be justified at any cost?
(a) $\dfrac{a \times b}{c} = x \times y$  (b)  $\dfrac{a \times b}{x} = yz$  (c) $\dfrac{xy}{z} = ab$

maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

  1. Only a

  2. Only c

  3. All the three

  4. Only b and c

Show answer
Correct Option: D
Explanation:
Use the rules that 
(a) product of two odd or even numbers are odd and even respectively.
(b) the quotient of the odd number $ \div $ even number or vice-versa may or may not be strictly odd or even.

a. $ ab $ must be a even number and $ \dfrac{ab}{c} $ may be even number or a fraction or odd number and $ xy $ must be a odd number
therefore it can be justified in some case.

b. $ \dfrac { ab }{ x } $ is always a fraction or even number and $ yz $ is always a odd number
therefore it cannot be justified in any case

c.$ \dfrac { xy }{ z } $ maybe odd number or fraction but $ ab $ will always be even.
therefore it cannot be justified.

Relationships given in b and c can't be justified at any cost.

If $f(x)=x^{2}+6x+c$, where $'c'$ is an integer, then $f(0)+f(-1)$ is

maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

  1. an even integer

  2. an odd integer always disable by $3$

  3. an odd integer not divisible by $3$

  4. an odd integer may or not be divisible by $3$

Show answer
Correct Option: D
Explanation:

$f(0) = c$

$f(-1) = c-5$

$f(0)+f(-1) = 2c-5 = 2(c-3) + 1$

As the above is of the form $2k+1$, it is always odd.

For $c=3$, the above is not divisible by 3 but for $c=4$, it is. Therefore, it may or may not be divisible by 3.

If $P$ is an integer between $0$ and $9,R-P=16229$ and $R$ divisible by $11$, then find the value of $\dfrac {P+R-1}{3}$

maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

  1. $5014$

  2. $4514$

  3. $5414$

  4. $5114$

Show answer
Correct Option: C
Explanation:
$ R-P = 16229 $
P be b/w $0\& 9 $
and R is divisible by 11
So, $ R = \dfrac{16229+P}{11} $
(and Reminder = 0)
So, $ \Rightarrow (\dfrac{11+P}{11}) $ so $ P = 7 $
and $ R = 16229+7 $
$ = 16236 $
So $ \dfrac{P+R+1}{3} $
$ \Rightarrow \dfrac{16236+7-1}{3} $
$ \Rightarrow \dfrac{16242}{3} = 5414 $ 
Option C is correct 

Difference of squares of two odd integers is always divisible by ?

maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

  1. 3

  2. 5

  3. 16

  4. 8

Show answer
Correct Option: D
Explanation:
Let two consecutive odd integers be $2k +1$ and $2k+3$, where $k$ is any integer
Difference$={\left(2k+3\right)}^{2}-{\left(2k+1\right)}^{2}$
$=4{k}^{2}+9+12k-4{k}^{2}-1-4k=8k$
Hence, the difference is always divisible by $8$.

Consider $n={21}^{52}$, then

maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

  1. number of even divisors of $n$ is $704$

  2. number of odd divisors of $n$ is $2809$

  3. last two digits of $n$ is $41$

  4. number of even divisors of $n$ which are multiple of $9$ is $2705$

Show answer
Correct Option: B,C
Explanation:

Let,we have


$n = {21^{52}}$

can be written as $n = {\left( {7 \times 3} \right)^{52}}$

$n = {7^{52}}{.3^{52}}$

We know, no. of total divisors of any number

$k = {p^m}.{q^n}$

Total divisors$=(m+1)\,(n+1)$

so, similary here

odd divisors$=(52+1)(52+1)=2809$

Hence the option $(B)$ is correct

But again 

For last two digit

$n = {21^{52}} = {\left( {20 + 1} \right)^{52}}$

${\left( {20 + 1} \right)^{52}}{ = ^{52}}{C _1}{\left( {20} \right)^{52}} + .....{ + ^{52}}{C _{51}}{\left( {20} \right)^1}{ + ^{52}}{C _{52}}{\left( {20} \right)^0}$

For last two digit we notice last two terms 

$=^{52}{C _{51}}\left( {20} \right) + 1$

$ = 52 \times 20 \times 1$

$=1041$

$1041$ has last two digit is $41$

so, option $(C)$ is also correct

Hence both the option $(B)$ and $(C)$ are correct.

The smallest odd number formed by using the digits $1,0,3,4$ and $5$ is

maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

  1. $10345$

  2. $10453$

  3. $10543$

  4. $10534$

Show answer
Correct Option: A
Explanation:

The smallest odd number using digits $1,0,3,4,5$


$\rightarrow $ We have five digits and we have to make smallest five digit odd numbers.


$\rightarrow$ So, the number cannot start with $0$

$\rightarrow$ For the smallest it should be start with $1$

$\rightarrow$ and second space should be $0$

    $1\\ \overline { 1st } $  $0\\ \overline { 2nd }$  $\;\\ \overline { 3rd } $  $\;\\ \overline { 4rt } $  $\;\\ \overline { 5th } $

$\rightarrow$ Now two space are filled and $3$ are left.

$\rightarrow$ For smallest third place for should be $3$ 

          $\underline { 1 } \underline { 0 } \underline { 3 } \underline {  } \underline {  } $

$\rightarrow $ Now two places are left for and no. should be odd so, last digit should be $5$

So, the number $=10345.$

The integer just below $(\sqrt{53}+7)^{11}-2\times 7^{11}$ is 

maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

  1. Divisible by exactly $4$ primes factors

  2. Divisible by exactly $3$ primes factors

  3. is divisible by $7$

  4. has $53$ as its only two digit prime factor

Show answer
Correct Option: A

Total number of four digit odd numbers that can be formed using $0,1,2,3,5,7$ are

maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

  1. $192$

  2. $375$

  3. $400$

  4. $720$

Show answer
Correct Option: D
Explanation:

we have the number $0,1,2,3,5,7$

Now the digit should be odd and hence last digit should be 
filled with an odd number 
$ \Rightarrow $ Number of way to filled last number $= 4$ $({\text{i}}{\text{.e}}{\text{. }}1,3,5,7)$
$ \Rightarrow $ Number of way to filled third digit  $= 6$ $({\text{i}}{\text{.e}}{\text{. 0,}}1,2,3,5,7)$
$ \Rightarrow $ Number of way to filled second digit $= 6$ $({\text{i}}{\text{.e}}{\text{. 0,}}1,2,3,5,7)$
$ \Rightarrow $ Number of way to filled first digit $= 5$ $({\text{i}}{\text{.e}.}1,2,3,5,7)$
$ \Rightarrow $ Total 4digit number $=4\times6\times6\times5$
$= 720$
hence,
Opton $D$ is correct answer.